Problem 50 - College Algebra

Question

If \(f(x)=x^{5},\) find \(\frac{f(x+h)-f(x)}{h}\) and simplify.

Step-by-Step Solution

Answer

The simplified form of the given expression \(\frac{f(x+h)-f(x)}{h}\) for the function \(f(x) = x^{5}\) is \(5x^{4} + 10x^{3}h + 10x^{2}h^{2} + 5xh^{3} + h^{4}\).

Step 1: Apply the Formula

Substitute \(f(x) = x^{5}\) in the formula \(\frac{f(x+h)-f(x)}{h}\) to get \(\frac{(x+h)^{5}-x^{5}}{h}\).

Step 2: Simplify the Numerator

Simplify the numerator using the binomial expansion theorem. \( (x+h)^{5} = x^{5} + 5x^{4}h + 10x^{3}h^{2} + 10x^{2}h^{3} + 5xh^{4} + h^{5}\). Then, \( \frac{(x+h)^{5}-x^{5}}{h} = \frac{x^{5}+ 5x^{4}h + 10x^{3}h^{2} + 10x^{2}h^{3} + 5xh^{4} + h^{5} - x^{5}}{h}\). This simplifies to \( \frac{5x^{4}h + 10x^{3}h^{2} + 10x^{2}h^{3} + 5xh^{4} + h^{5}}{h}\).

Step 3: Divide by h

Now divide every term in the numerator with \(h\) to get the final simplified expression. This results in \(5x^{4} + 10x^{3}h + 10x^{2}h^{2} + 5xh^{3} + h^{4}\).

Key Concepts

Function Simplification

Simplifying functions is a process that makes complicated mathematical expressions easier to work with. When dealing with functions like polynomials, simplification is essential for understanding their behavior and calculating derivatives.
One of the key simplification strategies involves substituting and reducing expressions. For instance, when you have an expression like \((x+h)^5 - x^5\), you can use the Binomial Theorem to expand \((x+h)^5\) and then subtract \(x^5\), which helps in simplifying the original function.
Another critical step is factoring and dividing by common terms to further reduce the expression. After simplifying, you're left with a more manageable function that retains the original function's properties, making it easier to apply calculus concepts like taking limits or derivatives.

• Substitute expressions as needed.
• Use algebraic expansion techniques.
• Divide by common terms to simplify.

This helps in reducing the workload when working through problems and provides a clearer pathway to the answer.

Binomial Theorem

The Binomial Theorem is a powerful algebraic tool used for expanding expressions raised to any power. It tells us how to expand and simplify expressions like \((x+h)^5\), which will appear when working with difference quotients or polynomial functions in calculus.
The theorem is defined by the formula:\[(x+y)^n = \sum_{k=0}^{n} \binom{n}{k} x^{n-k} y^k\]
Here, \(\binom{n}{k}\) is known as the binomial coefficient and can be calculated using factorials. In our example, \((x+h)^5\) expands to \(x^5 + 5x^4h + 10x^3h^2 + 10x^2h^3 + 5xh^4 + h^5\).

• This theorem expresses each term in the expansion as a combination of binomial coefficients and powers of the individual terms in the binomial expression.
• It's especially useful in calculus for simplifying the terms of a polynomial before finding the derivative.
• Knowing this theorem makes it easier to handle polynomials of higher degrees when simplifying difference quotients.

Being familiar with the Binomial Theorem saves time and effort in problems involving polynomial expansions.

Calculus

Calculus is a branch of mathematics focused on change and motion; it involves two main concepts: derivatives and integrals. For difference quotients, calculus helps find the derivative of a function, which represents the rate of change.
At its core, calculus uses the limit process to calculate derivatives by examining the slope of a function at any given point. The difference quotient provides a framework to understand this process: \[\frac{f(x+h)-f(x)}{h}\] as \( h \rightarrow 0 \).
This expression estimates how much \( f(x) \) changes as \( x \) changes by a small amount \( h \). By simplifying the difference quotient using algebraic techniques (like function simplification and the binomial theorem), we can compute the derivative more easily.

• Provides tools to study functions and their behavior.
• Focuses on the rate and pattern of change.
• Broadens the application to real-world problems like physics and engineering.

Ultimately, calculus extends the ideas learned through simplification and algebra to more complex systems, allowing deeper insights into dynamic phenomena.